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Homework Help: Forced Oscillation Q

  1. Feb 21, 2008 #1


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    [SOLVED] Forced Oscillation Q

    1. The problem statement, all variables and given/known data

    A 2.00kg object attatched to a spring moves without friction and is driven by an external force given by

    [tex]F= (3.00N)sin(2 \pi t) [/tex]

    The force constant of the spring is 20N/m. Determine

    a) period
    b) amplitude of motion

    2. Relevant equations

    [tex]T= 2 \pi / \omega [/tex]

    [tex]A= (F_o /m)/ \sqrt{ (\omega^2- \omega_o^2)^2 + (b \omega /m)^2} [/tex]

    3. The attempt at a solution

    a) [tex]T= 2 \pi / \omega [/tex]
    [tex]T= 2 \pi/ 2 \pi = 1 s [/tex]

    b) Um..[tex]A= (F_o /m)/ \sqrt{ (\omega^2- \omega_o^2)^2 + (b \omega /m)^2} [/tex]

    however I'm not sure as to what the

    [tex]F= (3.00N)sin(2 \pi t) [/tex]

    [tex]F_o= 3.00N [/tex]
    [tex] \omega = 2 \pi [/tex]

    I'm not sure as to what that omega in the given equation is..is it [tex]\omega_o [/tex] or just the [tex] \omega [/tex] ?

    b= 0 so that cancels out....

    Thanks alot
  2. jcsd
  3. Feb 21, 2008 #2


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    wo is the angular frequency of the driving force and w is the natural frequency of the oscillating spring. When w = wo the amplitude of the forced oscillation is maximum and that is the condition for the resonance.
  4. Feb 21, 2008 #3


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    but I was curious to know which one was given in the equation as I know it's written in my book that it is [tex] \omega [/tex] but I don't know which one. And after I know which one it is how do I find the other one since I want to find amplitude.
  5. Feb 21, 2008 #4


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    Actually the normal convention is the opposite way around, [tex]\omega_0[/tex] being the natural frequency.

    The natural frequency [tex]\omega_0[/tex] is equal to [tex]\sqrt{k/m}[/tex], where "k" is the spring constant.

    BTW. Obviously you're using a cookie cutter approach of substituting into "Relevant" equations so I'm guessing that a first principle appraoch of solving the systems differential equation is beyond the scope of your current course. I should however point out that the way you are solving this system is fundementally flawed in that with a truely frictionless systems the natural response cannot be ingored for any value of time t. Your "relevent equations" are only finding the particular solution and ingoring the homogenious solution which depends upon the initial conditions.

    Actually your approach would however be valid as a good approximation if your system is presented as a near frictionless system in steady state (note that a true frictionless system never actually reaches steady state).

    BTW. The truely relevent equation for this sytem is the DE: [tex]m \,\,d^2x/dt^2 + k x= A sin(wt)[/tex]
    Last edited: Feb 21, 2008
  6. Feb 22, 2008 #5


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    b) amplitude of motion

    [tex]A= (F_o /m)/ \sqrt{ (\omega^2- \omega_o^2)^2 + (b \omega /m)^2} [/tex]

    however I'm not sure as to

    b= 0 so that cancels out.... ====> this IS supposed to cancel out right??

    I can't figure the correct answer out....

    [tex]\omega = \omega_o[/tex] thus they SHOULD cancel out thus = 0
    However I said that since the damping was small then shouldn't (frictionless) b= 0 as well???

    But if this above is true then the bottom of the Amplitude eqzn would cancel out since it would = 0 + 0???

    Then in the end wouldn't the equation be just

    [tex]A= (F_o /m)/ \sqrt{ (\omega^2- \omega_o^2)^2 + (b \omega /m)^2} [/tex]

    [tex] A= F_o/m [/tex]

    (after cancelling out everything?)
    A= 3/2= 1.5 => wrong....

    Book answer is 5.09cm

    Okay so then I think that I actually plug in for omega's and b is I assume 0 since damping is nonexistant(frictionless)

    but after plugging in

    [tex]\omega= sqrt{ k/m} = sqrt {20.0N/m / 2.00kg}= 3.1623 [/tex]

    and [tex] \omega = 2 \pi [/tex] from equation given

    and what do I get??

    [tex]A= (F_o /m)/ \sqrt{ (\omega^2- \omega_o^2)^2 + (b \omega /m)^2} [/tex]

    [tex]A= (3.0N /2.00kg)/ \sqrt{ ((2 \pi )^2- (3.1623))^2 + (0)^2} [/tex]

    [tex]A= 1.5/ 5.429 = .27629m => 0.0027629m [/tex]

    Which is ALSO NOT the answer :uhh:

    Answer from book is 5.09cm

    Can someone please help me figure out why it isn't the answer the same as the book answer..

    Last edited: Feb 22, 2008
  7. Feb 22, 2008 #6


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    Natural frequency of the sping is given by wo^2 = k/m = 20/2 = 10
    Angular frequency of the driving force =w=2*pi. Substitute these values in the expression for amplitude( put b = 0 ), you will get the answer.
  8. Feb 22, 2008 #7


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    I got it THANKS rl.bhat :smile:
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